375.00 mL of a 0.010 M solution of NaCl are condensed down to 167.00 mL. Assumin

Question

375.00 mL of a 0.010 M solution of NaCl are condensed down to 167.00 mL. Assuming the solution does not become saturated during this process, what is the final concentration of NaCl?

Explanation / Answer

First of all let me make a correction. Sorry for the mistake. I was sleepy while answering the questions. It is related with the question 9.59. since ppm and ppb represents the same amount of concentration, it must the give same number. Therefore, while coverting ppm to ppb, in other words; parts / million ----> parts / billion, the numerator inceases by a factor of 1000. Therefore the numerator should also increase by a factor of 1000. So, the correct conversion will be ppb = ppm x 1000 not ppb = ppm / 1000. The correct answer is 0.050 ppm /x1000 = 50 ppb 9.65. As you remember, % w/v is the mass of the solute in g per 100 mL of the solution. Therefore, 0.40% (w/v) solution of nalorphine contains 0.40 g nalorpine in 100 mL of solution. 1.5 mg = 0.0015 g Volume of solution containing 0.0015 g nalorphine; 100 mL x ( 0.0015 g / 0.40 g) = 0.375 mL 9.69. MixVi = MfxVf This is a very useful formula to be used in either to DILUTE ot to CONCENTRATE the solutions.Subscript (i) denotes the initial state and bubscript (f) denotes the final state. You can use this this formula to calculate either molarity or volume. In case of volume; - If you dilute the solution by adding water, Mf > Mi, therefore the amount of water to be added will be Mf - Mi. - If you concentrate the solution by evaporating water, Mi > Mf, therefore the amount of water to be evaporated will be Mi - Mf. Also you don't need to convert mL to L. If we apply this to your question; MixVi = MfxVf (0.500 M)(100.0 mL) = (0.150 M) Vf Vf = (0.500 M)(100.0 mL) / (0.150 M) = 333.33 mL Water to be added : Vf - Vi = 333.33 - 100.0 = 233.33 mL 9.71. You can drive a similar formula; [(w/v)%]i x Vi = [(w/v)%]f x Vf First let's prove that this formula is correct. For this we will solve the problem without using formula: 50 mL of 27 (w/v)% solution contains 50 x 27/100 = 13.5 g NaCl. When the solution is diluted the amount of the solute does not change. Now we have 13.5 g NaCl in 420 mL of solution. (w/v)% of the new solution : (13.5 g / 420 mL) x 100 = 3.21 (w/v)% Now use the formula; [(w/v)%]i x Vi = [(w/v)%]f x Vf (27 %)(50) = [(w/v)%]f (420) [(w/v)%]f = (27 %)(50) / 420 = 3.21 (w/v)% 9.73. Use the first equation: MixVi = MfxVf (0.100 M)(Vi) = (0.0500 M)(750.0 mL) Vi = (0.0500 M)(750.0 mL) / (0.100 M) = 375 mL Take 375 mL 0.100 M NaHCO3 and dilute it to 750.0 mL by adding (750 - 375 =) 375 mL of water. Now the other questions are rather complicated. 9.77. Equivalent or equivalent weight (Eq) is a unit of relative amount of substance used in chemistry. One equivalent weight of an element, compound, or ion is the weight in grams of that substance which would react with or replace one gram of hydrogen. Since one gram of hydrogen is very nearly equal to one mole and since hydrogen has one electron free to react with other substances, 1 Eq of a substance is effectively equal to one mole divided by the valence of the substance (the number of electrons the substance would engage in participating in the reaction). In practice, this is a large unit and measurements are more likely to be in milliequivalents (mEq or meq). According to this explanation, since the valence (oxidation number) of the ions are considered in the calculation of the mEq, we will only add them. Total anion concentration; 5.0 + 12.0 + 2.0 = 19.0 mEq/L. 9.79. 0.025 M SO4^2- ion means that 1 liter of solution contains 0.025 mole SO4^2- ion. Therefore, we need to convert "mole" to Eq. (or mole/L to Eq/L). For this if you consider the above explanation, Eq = mole / valence Eq = 0.025 / 2 = 0.0125 Eq. 1 Eq = 1000 mEq So, 0.025 M (mol/L) SO4^2- = 0.0125 Eq/L = 12.5 mEq/L Note: Whether its sign is positive or negative; - For monovalent ions, Eq = mole - For divalent ions; Eq = mole / 2 - For trivalent ions; Eq = mole/3 9.81. 3 mEq/L = 0.003 Eq/L Since the valence (charge) of Mg^2+ ion is 2. Mole = 2 x Eq = 2 x 0.003 = 0.006 mol 3 mEq/L = 0.006 mol/L 1 L (1000 mL) contains 0.006 mol 100 mL contains 0.0006 mol 1 mol Mg^2+ is 24 g (molar mass of Mg) mass of Mg^2+ = 0.0006 mol x 24 g/ mol = 0.0144 g = 14.4 mg

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